Brian stated in earlier posts that the sequence is based on some algorithm and that the algorithm works on binary numbers. Here is what I've found...
The numbers show up in groups of 4, we'll call them A, B, C, and D.
A starts at 1.
To get B from A we shift all the 1's that are left of the right-most 0 to the far left.
To get C from B we tack on a 1 to the far left and a 1 to the far right.
To get D from C we chop off everything to the left of the central bit.
To get A' from A we tack on a 10 to the far left and a 11 to the far right. Then we start back at getting B' from A'...
The result is as follows with the binary number in the center and the decimal equivalent to the far right:
A = 1 = 1
B = 1 = 1
C = 111 = 7
D = 11 = 3
A' = 10111 = 23
B' = 10111 = 23
C' = 1101111 = 111
D' = 1111 = 15
A'' = 101011111 = 351
B'' = 110011111 = 415
C'' = 11100111111 = 1855
D'' = 111111 = 63
A''' = 1010101111111 = 5503
B''' = 1110001111111 = 7295
C''' = 111100011111111 = 30975
D''' = 11111111 = 255
A'''' = ...
Well? Does that work?
|
Posted by Erik O.
on 2004-06-10 13:08:07 |